Overview
| Comment: | 202409 1st star |
|---|---|
| Downloads: | Tarball | ZIP archive |
| Timelines: | family | ancestors | descendants | both | trunk |
| Files: | files | file ages | folders |
| SHA3-256: |
868207236374b9594ad435b7a82432c0 |
| User & Date: | nnz on 2024-12-09 11:50:44.623 |
| Other Links: | manifest | tags |
Context
|
2024-12-09
| ||
| 12:24 | changed empty values to work with part two check-in: c93635c614 user: nnz tags: trunk | |
| 11:50 | 202409 1st star check-in: 8682072363 user: nnz tags: trunk | |
|
2024-12-08
| ||
| 17:11 | 202408 2nd star check-in: 9395df6993 user: nnz tags: trunk | |
Changes
Modified aoc2024.c
from [05505307b8]
to [3ba7100c1a].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | #include <ctype.h> #include <stdbool.h> #include <stddef.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include "aocdailies.h" #include "aocutils.h" /* === aoc202408 ======================================================= Oh! another square of text! Idea: for all points p with an antenna find all points q>p with a corresponding frequency. For each such pair calculate the 2 antinodes and add the resulting points (if not there already) to an array. The answer to Part One is the number of elements in the array | > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 |
#include <ctype.h>
#include <stdbool.h>
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "aocdailies.h"
#include "aocutils.h"
/* === aoc202409 =======================================================
WOW! input consists of a 20,000 long string of numbers!
===================================================================== */
void aoc202409(char *data, size_t len) {
(void)len; // unused argument
int *disk = malloc(512 * sizeof *disk);
size_t rdisk = 512;
size_t ndisk = 0;
int fileid = 0;
while (*data) {
for (int k = 0; k < *data - '0'; k++) {
if (ndisk == rdisk) {
// assume it works
rdisk = (13*rdisk)/8;
disk = realloc(disk, rdisk * sizeof *disk);
}
disk[ndisk++] = fileid;
}
fileid++;
data++;
if (*data) {
for (int k = 0; k < *data - '0'; k++) {
if (ndisk == rdisk) {
// assume it works
rdisk = (13*rdisk)/8;
disk = realloc(disk, rdisk * sizeof *disk);
}
disk[ndisk++] = -1;
}
data++;
}
}
int *left = disk, *right = disk + ndisk;
for (;;) {
while (*left != -1) left++;
if (left >= right) break;
while (right[-1] == -1) right--;
// swap *left and right[-1]
int tmp = *--right;
*right = *left;
*left++ = tmp;
}
unsigned long long chksum = 0;
size_t index = 0;
while (disk[index] >= 0) {
chksum += (size_t)disk[index] * index;
index++;
}
printf("%llu\n", chksum);
free(disk);
}
/* === aoc202408 =======================================================
Oh! another square of text!
Idea: for all points p with an antenna find all points q>p with
a corresponding frequency. For each such pair calculate the 2 antinodes
and add the resulting points (if not there already) to an array.
The answer to Part One is the number of elements in the array
|
| ︙ | ︙ |
Modified aocdailies.c
from [fb8ca294f1]
to [53d57bd116].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
#include <stddef.h>
#include "aocdailies.h"
aocfunc *aocselect(unsigned y, unsigned d) {
aocfunc *p;
switch (y * 100 + d) {
default: p = NULL; break;
// YYYYdd ==> aocYYYYdd
case 202408: p = aoc202408; break;
case 202407: p = aoc202407; break;
case 202406: p = aoc202406; break;
case 202405: p = aoc202405; break;
case 202404: p = aoc202404; break;
case 202403: p = aoc202403; break;
case 202402: p = aoc202402; break;
| > | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 |
#include <stddef.h>
#include "aocdailies.h"
aocfunc *aocselect(unsigned y, unsigned d) {
aocfunc *p;
switch (y * 100 + d) {
default: p = NULL; break;
// YYYYdd ==> aocYYYYdd
case 202409: p = aoc202409; break;
case 202408: p = aoc202408; break;
case 202407: p = aoc202407; break;
case 202406: p = aoc202406; break;
case 202405: p = aoc202405; break;
case 202404: p = aoc202404; break;
case 202403: p = aoc202403; break;
case 202402: p = aoc202402; break;
|
| ︙ | ︙ |
Modified aocdailies.h
from [114faad3c2]
to [e29d12fccf].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | #ifndef AOCDAILIES_H_INCLUDED #define AOCDAILIES_H_INCLUDED #include <stddef.h> typedef void aocfunc(char *, size_t); aocfunc *aocselect(unsigned, unsigned); aocfunc aoc202408; aocfunc aoc202407; aocfunc aoc202406; aocfunc aoc202405; aocfunc aoc202404; aocfunc aoc202403; aocfunc aoc202402; | > | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #ifndef AOCDAILIES_H_INCLUDED #define AOCDAILIES_H_INCLUDED #include <stddef.h> typedef void aocfunc(char *, size_t); aocfunc *aocselect(unsigned, unsigned); aocfunc aoc202409; aocfunc aoc202408; aocfunc aoc202407; aocfunc aoc202406; aocfunc aoc202405; aocfunc aoc202404; aocfunc aoc202403; aocfunc aoc202402; |
| ︙ | ︙ |